Download Test #4 - Yeah, math, whatever.

Practice Test #4 MAT 120
(1) For the experiment 'toss the same coin 3 times', let the variable X correspond to the
number of tails in the three tosses. Construct a discrete probability distribution for the
experiment.
(2) The probabilities of a machine manufacturing 0,1, 2, or 3 books are .45, .3, .15, and
.10, respectively. Construct a discrete probability distribution for this data.
(3) Tell whether or not the example is a discrete probability distribution. If not, state why.
(a)
X
|
P(X) |
(b)
X
|
P(X) |
(c)
X
|
P(X) |
1
.3
2
.15
3
.4
4
.2
5
.1
1
1
5
2
1
5
3
1
5
4
1
5
5
1
5
1
.2
2
1.1
3
-.9
4
-.5
5
.1
(4) A study conducted by a TV station showed the number of televisions per household
and the corresponding probabilities for each. Find the mean(/expected value).
# Televisions X |
1
2
3
4
P(X)
| .32
.51
.12
.05
In numbers 5 and 6, use the rules concerning the binomial distribution.
(5) If a coin is tossed 7 times, what is the probability that exactly 4 heads will be thrown?
(6) In a survey, 75 percent of people polled admitted they were clueless when it comes to
anything important. If 9 people from the survey are picked at random, what is the
probability that exactly 4 of them will admit they're clueless?
(7) Determine the area under the standard normal curve that lies
(a) to the left of z  2.31
(b) to the right of z  .57
(c) between z  2.31 and z  .57
(8) A big exam is given to high school students across the country. The scores are
normally distributed, with mean score 1025, and a standard deviation of 100 points. What
percent of students would we expect to find scoring in the following ranges?
(a) 900-1200?
(b) Less than 900?
(c) Greater than 1200?
(9) The average lifetime of a machine widget is 720 days, with a standard deviation of 30
days. What are the minimum and maximum lifespans that put widgets in the 86th
percentile? Assume the variable is normally distributed.
(10) The mean score on a dexterity test for 120 year-olds is 30. The standard deviation is
5. If a psychologist administers the test to a class of 22 students, find the probability that
the mean of the sample will be between 27 and 31. Assume the variable is normally
distributed. (careful – note that the question asks about the 'probability of the mean').
Solutions:
(1) The number of tails can be 0, 1, 2, or 3. You may want to use a tree diagram to list all
of the outcomes. {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
#TAILS X |
0
1
2
3
1
3
3
1
P(X)
|
8
8
8
8
(2)
#BOOKS X
|
0
1
2
3
P(X)
| .45
.3
.15
.1
(3) (a) no;
.3 + .15 + .4 + .2 + .1 = 1.15. The probabilities cannot sum to more than
one.
(b) yes; all of the probabilities are between 0 and 1, and they add up to 1.
(c) No; one of the probabilities is negative, and another is greater than 1
(4) formulas:    X * P( X ),
the mean is:
  1*.32  2*.51  3*.12  4*.05  1.9,
(5) Formula: n Ck ( p) k (q) n k , q  1  p
define: 'success' as head, with probability p =
1
1 1
, q = 1 - = , we want the probability
2
2 2
of 4 'successes' in 7 'trials'.
4
74
1 1
7 C4 
    .27
2 2
on the calculator, 2nd VARS and scroll down to binompdf, then enter the number of
trials, probability of success, and the number of successes:
(6) Paradoxically, let's let clueless mean 'success', and p = .75, so not clueless would be
'failure', with q = 1 - .75 = .25 So, we're looking for 4 successes in 9 trials:
C4 .75  .25 
4
9
94
 .04
on the calculator, 2nd VARS and scroll down to binompdf, then enter the number of
trials, probability of success, and the number of successes:
(7)
(a) On the TI-8x, hit 2nd VARS and choose normalcdf. Then enter the 'exponential
zero', the z-score -2.31, a 0, a 1 and hit enter:
(b) Do the above with z-score z = .57, then subtract that number from 1, thus using the
complement argument:
(c) You can use the same function from the calculator, but just feed the two z-scores
instead of the 'exponential zero':
(8)
(a) Find the z-scores:
900  1025 125
1200  1025 175
z900 

 1.25 , z1200 

 1.75
100
100
100
100
With either the table or the calculator, find the area between these two z-values:
(b) Subtract the corresponding z-score from .5:
Just find the area to the left of z = -1.25
(c) Find the area to the left of z = 1.75, and subtract it from 1:
.
(9) Either do a 'reverse lookup' on the value .86 in table 5 to find the corresponding zscore, or hit 2nd VARS and use invNorm – enter .86, 0, and 1:
Now we need to find the data value X that goes with z-value z = 1.08
X  z     , so Xmax = 1.08 * 30 + 720 = 752. I rounded down here.
(10) If the question asks for the probability of the mean this or that, then you need to use
the statistics concerning the mean from the Central Limit Theorem.
X 
z
is the z-score we use in such situations. The population mean is   30. We
/ n
need to find the probability that the sample mean X is between 27 and 31. Find the zscore for each of these numbers.
27  30
31  30
z
 2.814, z 
 .94,
5 / 22
5 / 22